$$\int\tan^2 xdx=\tan x-x+C$$
Work Step by Step
$$A=\int\tan^2 xdx$$ Remember that $\sec^2 x=1+\tan^2 x$. So, $\tan^2 x=\sec^2 x-1$ $$A=\int(\sec^2x-1)dx$$ Also, $\int\sec^2 x=\tan x+C$. Therefore, $$A=\tan x-x+C$$
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